Let \(f\) be entire and for every \(a\in \CC\), and write the Laurent expansion about \(a\) as \(f(z) = \sum_{k\geq 0} c_{k, a} (z-a)^k\). Suppose that for every \(a\), there exists some \(k\) such that \(c_{k, a} = 0\). Show that \(f\) is a polynomial.
Solution
Claim: \(f^{(n)} = 0\) for some \(n\).
Let \(Z_n \da \ts{a\in \CC \st c_{n, a} = 0}\) and \(Z = \Union_n Z_n\). Since \(c_{n,a}=f^{(n)}(a)/n!\), we have \(f^{(n)}(z) = 0\) on \(Z_n\).
By hypothesis, \(Z = \CC\), so not every \(Z_n\) can be countable. Pick any \(Z_n\) which is uncountable.
Write \(Z_n = \Union_{m} Z_{m, n}\) where \(Z_{m, n} \da \DD_m(0) \intersect Z_n\) as a union of increasing discs, then not every \(Z_{m, n}\) can be finite, so pick an infinite one.
Now the infinite set \(Z_{m, n}\) has a limit point in \(\CC\), since it lies in the compact closed disk \(\overline{\DD_m(0)}\), and \(f^{(n)}(z) = 0\) on \(Z_{m, n}\), so by the identity principle \(f^{(n)}(z) = 0\) on \(\CC\).
Since \(f^{(n)}\equiv0\), all Taylor coefficients of \(f\) at \(0\) of index at least \(n\) vanish, so \(f\) is a polynomial of degree less than \(n\).